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Question: Find the mode of \[10\] , \[12\], \[11\], \[10\], \[15\], \[20\],\[19\], \[21\], \[11\], \[9\], \[10...

Find the mode of 1010 , 1212, 1111, 1010, 1515, 2020,1919, 2121, 1111, 99, 1010.

Explanation

Solution

Here we will find the Mode of the given data by checking that number which is appearing most often in the series. For example, in the series 11, 22, 22, 33 and 44 the number 22 is appearing maximum times in respect of the rest numbers. So, the mode will be equals 22.

Complete step-by-step solution:
Step 1: First of all, we will check how many times a number is repeating itself in the series as shown below:
\RightarrowThe number 99 is repeated only one time.
\RightarrowThe number
1010 is repeated three times.
\RightarrowThe number 1111 is repeated two times.
\RightarrowThe number 1212 is repeated only one time.
\RightarrowThe number 1515 is repeated only one time.
\RightarrowThe number 1919 is repeated only one time.
\RightarrowThe number 2020 is repeated only one time.
\RightarrowThe number 2121 is repeated only one time.
Step 2: As we know that the mode of the series will be equal to that number that is repeating most often. So, in the above-given data the mode will be equals to as below:
Mode=10\Rightarrow {\text{Mode}} = 10 (because it is repeating three times)\because \left( {{\text{because it is repeating three times}}} \right)

The mode will be equal to 1010.

Note: Students should remember that if data is given with the class intervals with their respective frequency then the formula for calculating mode will be equals to as below:
Mode=l+(f1f02f1f0f2)h{\text{Mode}} = l + \left( {\dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right)h , where ll is the lower limit of the modal class, hh is the size of the class interval, f1{f_1} is known as the frequency of the modal class, f0{f_0} is known as the frequency of the class preceding the modal class, and f2{f_2} is known as the frequency of the class succeeding the modal class.